The circuit maximum voltage is around 25.2V so this is the supply im using, I wonder which one is the best way to get 3v3 with less components count and keeping it cost effective, space is a problem, smd components can be used.

PD: Its going to an ESP8266-12e, in datasheet maximum current is 170mA.

  • \$\begingroup\$ Output current? \$\endgroup\$
    – winny
    Mar 29, 2017 at 17:25
  • \$\begingroup\$ best in what sense? just cost and space? no consideration on efficiency at all? \$\endgroup\$ Mar 29, 2017 at 17:27
  • \$\begingroup\$ @winny, Its going to an ESP8266-12e, in datasheet maximum is 170mA. \$\endgroup\$ Mar 29, 2017 at 17:27
  • \$\begingroup\$ @user3528438, if having a good efficiency is not too much asking would be great, but space is really important. \$\endgroup\$ Mar 29, 2017 at 17:31
  • \$\begingroup\$ Edit your question and and that vital information. It's a bit if short duty cycle but most off the shelf DCDC converter will do it for you. Go to TI webench and enter your parameters. \$\endgroup\$
    – winny
    Mar 29, 2017 at 17:32

2 Answers 2


I think your best option would be to use a switch mode power supply. While they are more complex than using a linear regulator they can be made fairly small.

The problem with a simpler linear regulator is not just that they are inefficient, it is that the loss is turned into heat. At 25.2V the regulator would have to drop 21.9V to reach 3.3V. If the ESP-8266 draws around 170mA it will result in 3.723 Watts of heat.

You could manage this with a heat sink but this will take up space and radiate heat into your device.

A switch mode power supply will only produce small amounts of heat because it is far more efficient.


You've listed the max voltage, but you really need a voltage range. I'll just answer about regulators and if someone wants to go beyond my answer, they can.

1) If you have a specific voltage range (e.g 22.2V-25.2V) you can use a linear regulator with a resistor in series. How do you size these? It's just a voltage divider. This is very low efficiency, but depending on your current draw/application, it might be what you need.

V = I*R => P = I^2*R and P = I*V What's the current of your circuit? That's your I through the regulator. V_drop = I_reg * R_series (V_drop < (V_min - V_reg)) P_res = I_reg^2*R_series P_reg = I_reg*(V_bat-V_drop-V_reg) Make sure you do not dissipate too much power

You have a voltage range of Vbat = [22.2V; 25.2V] You have a regulator voltage of 3.3V. You have a 3.3V circuit that draws 170mA.

We want V_drop < (V_min - V_reg) V_drop < (22.2-3.3) V_drop < 18.9 (let's say V_drop = 18 to be safe) V_drop = 18

R = V/I = 18/0.17 = 106 Ohm. P = I^2*R = 106*.17*.17 = 3W (you could use a bunch of parallel or series resistors for this)

P_reg = (V_bat - V_drop- V_reg)*I P_reg = (25.2 - 18 - 3.3)*0.17 = 0.663W

Don't really think this is practical....

2) Use a switching regulator. These regulators can take a variety of inputs with a high efficiency and put out a variety of outputs. Be sure to look at the datasheet. The issue with switchers is that they cause higher EMI and you probably want some more filtering.

Here's a decent 1A switcher. Be sure to read the datasheet. I would recommend using a pi-filter on the input to the 3.3V with a cutoff frequency greater than 260kHz with this.



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